$\begingroup$How to show that X−A is a rank 2 matrix??$\endgroup$
– Debasish JanaSep 20 '18 at 15:05

$\begingroup$Welcome to MSE. It is more likely that you will get responses when you share your efforts.$\endgroup$
– Ahmad BazziSep 20 '18 at 15:06

$\begingroup$$X-A= ab^T A + Aba^T+ ab^T A ba^T$ ... Hence each of the terms are having rank 1. so the sum of all the terms can have rank $\leq 3$.. But I am not getting how it can be exactly of rank 2..$\endgroup$
– Debasish JanaSep 20 '18 at 15:12

2 Answers
2

We can write $X - A$ as
$$
X - A = ab^T[A + Aba^T] + Aba^T = ab^TA[I + ba^T] + Aba^T
$$
Noting that $\operatorname{rank}(PQ) \leq \min\{\operatorname{rank}(P),\operatorname{rank}(Q)\}$, we can see that each term has rank at most $1$, which means that $X - A$ has rank at most $2$.

It now remains to be shown that the rank is not $1$ or $0$.

First, we must show that the first matrix in the sum is non-zero. That is, we wish to show that the product
$$
b^T A[I + ba^T]
$$
is not the zero matrix. To that end, we note that
$$
(b^T A[I + ba^T])b = b^T A[b + b(a^Tb)] = (1 + a^Tb)(b^TAb)b \neq 0
$$
From there, it suffices to note that the first term has the span of $a$ as its column space, while the second term has the span of $Ab$ as its column space. Thus, the two column spaces are distinct and one-dimensional. It follows that the sum of the two non-zero rank $1$ matrices must have rank $2$.

$(X-A)$ is spanned by two vectors only

We can solve this using projector matrices, i.e. matrices of the form
\begin{equation}
P = Z(Z^TZ)^{-1}Z^T
\end{equation}
The number of columns of $X$ will determine the rank of the matrix $X -A $. If we stack in $Z$,
\begin{equation}
Z =
\begin{bmatrix}
a & Ab
\end{bmatrix}
\end{equation}
We are sure that $Z$ is full column rank because $a$ is not parallel to $Ab$, Hence it is easy to see that
\begin{equation}
Pa = a \tag{1}
\end{equation}
and
\begin{equation}
PAb = Ab \tag{2}
\end{equation}
Hence
$$P(X-A) = P(ab^T A + Aba^T+ ab^T A ba^T)$$
which is
$$P(X-A) = Pab^T A + PAba^T+ Pab^T A ba^T$$
Using equations $(1,2)$, we get
$$P(X-A) = ab^T A + Aba^T+ ab^T A ba^T = X-A$$
Hence $X-A$ is spanned by two vectors, i.e. rank $2$.
Therefore $P$ spans a two dimensional space, which is the span of the column space of $X -A$. On the other hand, we can also show that
\begin{equation}
(I-P).(X-A) = 0
\end{equation}
where $I-P$ is of rank $n-2$.